Hamiltonian Equations

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  • MechanicsPhysics 151

    Lecture 18Hamiltonian Equations of Motion

    (Chapter 8)

  • Whats Ahead

    We are starting Hamiltonian formalismHamiltonian equation Today and 11/26Canonical transformation 12/3, 12/5, 12/10Close link to non-relativistic QM

    May not cover Hamilton-Jacobi theoryCute but not very relevant

    What shall we do in the last 2 lectures?Classical chaos?Perturbation theory?Classical field theory?Send me e-mail if you have preference!

  • Hamiltonian Formalism

    Newtonian Lagrangian HamiltonianDescribe same physics and produce same resultsDifference is in the viewpoints

    Symmetries and invariance more apparentFlexibility of coordinate transformation

    Hamiltonian formalism linked to the development ofHamilton-Jacobi theoryClassical perturbation theoryQuantum mechanicsStatistical mechanics

  • Lagrange Hamilton

    Lagranges equations for n coordinates

    n equations 2n initial conditions

    Can we do with 1st-order differential equations?Yes, but youll need 2n equationsWe keep and replace with something similar

    We take the conjugate momenta

    0i i

    d L Ldt q q

    =

    1, ,i n= 2nd-order differential

    equation of n variables

    ( 0)iq t = ( 0)iq t =

    iq iq( , , )j j

    ii

    L q q tp

    q

  • Configuration Space

    We considered as a point in an n-dim. spaceCalled configuration spaceMotion of the system A curve in the config space

    When we take variations,we consider and asindependent variables

    i.e., we have 2n independent variables in n-dim. spaceIsnt it more natural to consider the motion in 2n-dim space?

    1( , , )nq q

    ( )i iq q t=

    iq iq

  • Phase Space

    Consider coordinates and momenta as independentState of the system is given byConsider it a point in the 2n-dimensional phase space

    We are switching theindependent variables

    A bit of mathematical trickis needed to do this

    1 1( , , , , , )n nq q p p

    ( )( )

    i i

    i i

    q q tp p t

    =

    =( , , ) ( , , )i i i iq q t q p t

  • If and

    Legendre Transformation

    Start from a function of two variablesTotal derivative is

    Define and consider its total derivative

    i.e. g is a function of u and y

    ( , )f x y

    g f ux

    f fdf dx dy udx vdyx y

    = + +

    ( )dg df d ux udx vdy udx xdu vdy xdu= = + =

    g vy

    =

    g xu

    =

    f L= ( , ) ( , )x y q q=

    ( , ) ( , )L q q g p q L pq =

    This is what we need

  • Hamiltonian

    Define Hamiltonian:Total derivative is

    Lagranges equations say

    This must be equivalent to

    ( , , ) ( , , )i iH q p t q p L q q t=

    i i i i i ii i

    L L LdH p dq q dp dq dq dtq q t

    = +

    Opposite sign from Legendre transf.

    ii

    L pq

    =

    i i i iLdH q dp p dq dtt

    =

    i ii i

    H H HdH dp dq dtp q t

    = + +

    Putting them together gives

  • Hamiltons Equations

    We find and

    2n equations replacing the n Lagranges equations1st-order differential instead of 2nd-orderSymmetry between q and p is apparent

    There is nothing new We just rearranged equationsFirst equation links momentum to velocity

    This relation is given in Newtonian formalismSecond equation is equivalent to Newtons/Lagranges equations of motion

    ii

    H qp

    =

    ii

    H pq

    =

    H Lt t

    =

  • Particle under Hookes law force F = kx

    Hamiltons equations

    Quick Example

    2 2

    2 2m kL x x=

    2 2

    22

    2 2

    2 2

    m kH xp L x x

    p k xm

    = = +

    = +

    Lp mxx

    = =

    Hp kxx

    = =

    H pxp m

    = =

    Replace withx pm

    Usual harmonic oscillator

  • Energy Function

    Definition of Hamiltonian is identical to the energy

    function

    Difference is subtle: H is a function of (q, p, t)

    This equals to the total energy ifLagrangian isConstraints are time-independent

    This makesForces are conservative

    This makes

    ( , , ) ( , , )ii

    Lh q q t q L q q tq

    =

    0 1 2( , ) ( , ) ( , )i j kL L q t L q t q L q t q q= + +

    2 ( , ) j kT L q t q q=

    0 ( )V L q=

    See Lecture 4, or Goldstein Section 2.7

  • Hamiltonian and Total Energy

    If the conditions make h to be total energy, we can skip calculating L and go directly to H

    For the particle under Hookes law force

    This works often, but not alwayswhen the coordinate system is time-dependent

    e.g., rotating (non-inertial) coordinate systemwhen the potential is velocity-dependent

    e.g., particle in an EM field

    22

    2 2p kH E T V xm

    = = + = +

    Lets look at this

  • Particle in EM Field

    For a particle in an EM field

    Wed be done if we were calculating h

    For H, we must rewrite it using

    2

    2 i i imL x q qA x= + We cant jump on H = E

    because of the last term, but

    i i ip mx qA= +2

    ( )2

    ii i i

    mxH mx qA x L q= + = + This is in fact E

    2( )( , )2

    i ii i

    p qAH x p qm

    = +

    i i ip mx qA= +

  • Particle in EM Field

    Hamiltons equations are

    Are they equivalent to the usual Lorentz force?Check this by eliminating pi

    2( )( , )2

    i ii i

    p qAH x p qm

    = +

    i ii

    i

    p qAHxp m

    = =

    j j j

    ii i i

    p qA AHp q qx m x x

    = =

    ( ) ji i ii i

    Ad mx qA qx qdt x x

    + =

    ( ) ( )i i id mv qE qdt

    = + v BA bit of work

  • Conservation of Hamiltonian

    Consider time-derivative of Hamiltonian

    H may or may not be total energyIf it is, this means energy conservationEven if it isnt, H is still a constant of motion

    ( , , )dH q p t H H Hq pdt q p t

    Hpq qpt

    = + +

    = + +

    Hamiltonian is conserved if it does not depend explicitly on t

  • Cyclic Coordinates

    A cyclic coordinate does not appear in LBy construction, it does not appear in H either

    Hamiltons equation says

    Exactly the same as in the Lagrangian formalism

    (H q , , ) (i ip t q p qL= , , )q t

    0Hpq

    = =

    Conjugate momentum of a

    cyclic coordinate is conserved

  • Cyclic Example

    Central force problem in 2 dimensions

    Cyclic variable drops off by itself

    2 2 2( ) ( )2mL r r V r= +

    rp mr=2p mr =

    22

    2

    22

    2

    1 ( )2

    1 ( )2

    r

    r

    pH p V rm r

    lp V rm r

    = + +

    = + +

    is cyclic constp l = =Hamiltons equations

    rprm

    =2

    3

    ( )r

    l V rpmr r

    =

  • Going Relativistic

    Practical approachFind a Hamiltonian that works

    Does it represent the total energy?

    Purist approachConstruct covariant Hamiltonian formalism

    For one particle in an EM field

    Dont expect miraclesFundamental difficulties remain the same

  • Practical Approach

    Start from the relativistic Lagrangian that works

    It does equal to the total energyHamiltons equations

    2 21 ( )L mc V= x

    21i

    ii

    mvLpv

    = =

    Did this last time

    2 2 2 4 ( )H h p c m c V= = + + x

    2

    2 2 2 4i i

    ii

    p c pHxp mp c m c

    = = =

    +i i

    i i

    H Vp Fx x

    = = =

  • Practical Approach w/ EM Field

    Consider a particle in an EM field

    Hamiltonian is still total energy

    Difference is in the momentum

    2

    2 2 2 2 2 4

    H m c q

    m v c m c q

    = +

    = + +

    Can be easily checked

    2 21 ( ) ( )L mc q q = + x v A

    i i ip m v qA= +2 2 2 4( )H q c m c q= + +p A

    Not the usual linear momentum!

  • Practical Approach w/ EM Field

    Consider H q

    It means that is a 4-vector,and so is

    This particular Hamiltonian + canonical momentum transforms as a 4-vector

    True only for well-defined 4-potential such as EM field

    2 2 2 4( )H q c m c q= + +p A

    2 2 2 2 4( ) ( )H q q c m c =p A constant

    ( , )H q c q c p A( , )H cp Similar to 4-momentum (E/c, p) of

    a relativistic particle

    Remember p here is not the linear momentum!

  • Purist Approach

    Covariant Lagrangian for a free particle

    We know that p0 is E/cWe also know that x0 is ct

    Generally true for any covariant LagrangianYou know the corresponding relationship in QM

    12 mu u

    =

    p muu

    = =

    2p p

    Hm

    =

    Energy is the conjugate momentum of time

  • Purist Approach

    Value of Hamiltonian is

    What is important is Hs dependence on p

    Hamiltons equations

    Time components are

    dx H pd p m

    = =

    0dp Hd x

    = =

    2

    2 2p p mcH

    m

    = =

    4-momentum conservation

    This is constant!

    ( )d ct E cd mc

    = =( ) 0d E cd

    = Energy definitionand conservation

  • Purist Approach w/ EM Field

    With EM field, Lagrangian becomes

    Hamiltons equations are

    A bit of work can turn them into

    12( , )x u mu u qu A

    = + p mu qA

    = +

    ( )( )2 2

    mu u p qA p qAH

    m

    = =

    dx H p qAd p m

    = =

    ( )p qAdp H Ad x m x

    = =

    du A Am q u Kd x x

    = = 4-force

  • EM Field and Hamiltonian

    In Hamiltonian formalism, EM field always modify the canonical momentum as

    A handy rule:

    Often used in relativistic QM to introduce EM interaction

    0Ap p qA = +